3y^2=-4y

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Solution for 3y^2=-4y equation:



3y^2=-4y
We move all terms to the left:
3y^2-(-4y)=0
We get rid of parentheses
3y^2+4y=0
a = 3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*3}=\frac{-8}{6} =-1+1/3 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*3}=\frac{0}{6} =0 $

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